Convergence/Divergence of the series $\sum_{n=1}^\infty\frac{(-1)^n\log n}{\sqrt n}$ According to the answers in my book, the series converges.
I know I should use Alternating Series Test and the limit of ln (n)/(sqrt n) approaches 0. 
And the derivative is negative after n=8, but the derivative is positive when 1<=n<=8, is that ok to use Alternating Series Test? I remember the condition is that it should decrease for all n>=1.
 A: It seems to be you mean the series $\;\displaystyle\sum\limits_{n=1}^\infty\frac{(-1)^n\log n}{\sqrt n}\;$ . Now, we have
$$f(x):=\frac{\log x}{\sqrt x}\implies f'(x)=\frac{\frac1{\sqrt x}-\frac{\log x}{2\sqrt x}}{x}=\frac{2-\log x}{2x\sqrt x}<0\iff\log x>2\iff x>e^2$$
and then the sequence $\;\left\{\,\frac{\log n}{\sqrt n}\,\right\}_{n\in\Bbb N}\;$ is mononotonic decreasing ( and clearly positive for all $\;n>1\;$) for $\;n>e^2\implies n\ge8\;$ , and thus we have here a Leibniz series which  converges (conditionally, though)
A: Not an answer, just a note.
We can actually find the closed form of this series, by considering a more general one:
$$f(s)=\sum_{n=1}^\infty (-1)^n \frac{\log n}{n^s}$$
Let's introduce an auxiliary function:
$$g(s)=\sum_{n=1}^\infty (-1)^{n+1} \frac{1}{n^s}=(1-2^{1-s}) \zeta(s)$$
This is true by definition of http://mathworld.wolfram.com/DirichletEtaFunction.html. Though, we can prove it independently as well.
On the other hand, we can see:
$$f(s)=\frac{d}{ds} g(s)$$
So we can immediately write:
$$f(s)=\frac{d}{ds}  \left[(1-2^{1-s}) \zeta(s) \right]=2^{1-s} \zeta(s) \log 2 +(1-2^{1-s}) \zeta'(s)$$
Derivative of zeta function is usually quite complicated and doesn't always have a closed form.
However for $s=1/2$ it's a special case which allows us to use the so called "reflection formula" to get the closed form, as can be seen from this answer: https://math.stackexchange.com/a/2495423/269624
$$\zeta'\left(\frac{1}{2}\right) = \frac{1}{2} \zeta\left(\frac{1}{2}\right) \left( \log \pi+\gamma+\frac{\pi}{2}+3\log 2 \right)$$
Which finally gives us:

$$f \left(\frac{1}{2}\right) =\frac{1}{2} \zeta\left(\frac{1}{2}\right) \left[2\sqrt{2} \log 2-(\sqrt{2}-1) \left( \log \pi+\gamma+\frac{\pi}{2}+3\log 2 \right) \right]$$

